USPAS notes Barletta,Spentzouris,Harms

1 Dipole errors and correction

1.1 Closed orbit change from magnetic dipole errors

In order for a particle to repeat the same orbit around an accelerator on every revolution, itsposition and angle must return to the same value when it returns to the same longitudinallocation in the ring. A machine is designed so that a particle initially on the reference orbit(with position x0 and angle x

′

0) will stay on that orbit. Denoting the transfer matrix of thethe ring as M , this means that

(

x0x

′

0

)

c.o.

=M

(

x0x

′

0

)

c.o.

The subscript c.o. indicates that the particle must be on the closed orbit for therelation to hold. A particle injected with a position or angle error will not return to itsinitial coordinates after a single revolution. It oscillates about the design trajectory, anddoes not undergo an integer number of oscillations per revolution. The phase space ellipsefor a particular location in the machine is defined by the machine parameters β and α anddoesn’t change. The errant particle will hit a different point on its phase space ellipse everytime it goes another turn around, but it is constrained to be somewhere on its particularellipse. On the other hand, the ideal particle is in the center of the phase space ellipse onevery revolution, returning to the same location in phase space every time.

There are sources of error in an accelerator that cause the closed orbit to deviate fromthe original designed trajectory. The entire beam envelope oscillates around the originaldesign orbit; there is a new closed orbit. On this new orbit there is still an ideal closedtrajectory; a particle on this particular trajectory will retrace its path turn after turn. Notethat when a beam is injected into a machine properly, the beam centroid is injected ontothe closed orbit with no error in position or angle. Then, the center of mass of the beamfollows the actual closed orbit (as opposed to the designed closed orbit). This orbit canthen be seen by a decent beam position monitoring system.

The orbit is determined by the magnetic fields through which the beam travels. So, thereare various types of errors that can be in a machine, dipole field errors, quadrupole fielderrors, and so on. Possible sources of dipole error include dipole magnets that have been

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’rolled’ in their placement, or quadrupoles with their centers misaligned to the referenceorbit. A dipole can also have a weaker or stronger ~B field due to errors in its construction.These errors cause an angular deflection of the beam, causing the closed orbit to oscillatearound the design orbit. A single steering error in some locations can cause a visible kinkin the closed orbit. It is desirable to analyze the effect of different types of errors on thebeam. A single dipole error gives an unwanted angular kick to the beam. So, as the closedorbit passes through the field error, the ideal particle’s position does not change, but itsangle changes according to the angular kick. The new closed orbit incorporates this kick.If the new closed orbit coordinates are x, x’; then for each revolution, the effect of the fieldsexperienced by design plus the kick must result in the same coordinates x, x’.

I

(

xx

′

)

c.o.

=M

(

xx

′

)

c.o.

+

(

0θ

)

where I is the identity matrix,

I =

(

1 00 1

)

Then,

(M − I)

(

xx

′

)

c.o.

=

(

0−θ

)

Now we have a relation between the strength of the error and the change in positionand angle of the closed orbit trajectory at the location of the error. Make this more explicitby writing M in terms of machine parameters;

M − I =

(

cos (2πν) + α sin (2πν)− 1 β sin (2πν)−γ sin (2πν) cos (2πν)− α sin (2πν)− 1

)

The two simultaneous equations with two unknowns given by the matrix equation are,

[cos (2πν) + α sin (2πν)− 1]x+ [β sin (2πν)]x′

= 0

[−γ sin (2πν)] x+ [cos (2πν)− α sin (2πν)− 1]x′

= −θ

The solutions are,

x =βθ

2

(

cos (πν)

sin (πν)

)

2

x′

=θ

2

(

sin (πν)− α cos (πν)

sin (πν)

)

What about the new closed orbit positions at some other location in the ring?

(

x2x

′

2

)

=M12

(

x1x

′

1

)

where

M12 =

β2

β1

[cos (ψ12) + α1 sin (ψ12)] (β1β2)1

2 sin (ψ12)

− 1+α1α2

(β1β2)1

2

sin (ψ12) +α1−α2

(β1β2)1

2

cos (ψ12)(

β1

β2

)1

2 cos (ψ12)− α2 sin (ψ12)

so that

x2 =θ√β1β2

2 sin (πν)cos (ψ12 − πν) (1)

Notice that Eq. 1 goes to infinity if the number of betatron oscillations per turn, ν,is an integer value. Dipole errors drive what is called an ’integer resonance’. Should thetune (ν) of a circular machine be set to an integer value, the presence of a dipole error(and there is always some dipole error present) will cause the betatron amplitude to growwithout bound.

When an accelerator is built, there are multiple errors causing a closed orbit distortion.The locations and number of errors is unknown! In the process of designing an accelerator,it is important to make an estimation of what these random errors do to the orbit, sothat an appropriate correction system can be put into place. The errors may be summed,assuming a random distribution.

x =

√βs

2 sin (πν)

Nerror∑

i

√

βiθi cos [(ψs − ψi)− πν]

Some of the errors will cancel each other since they are of opposite sign. Due to thevariation in the signs of the errors, it is more useful to estimate the mean square error ofthe beam centroid.

⟨

x2c.o.⟩

= βs1

4 sin2 (πν)

⟨

Nerror∑

i

βiθ2i cos

2 [(ψs − ψi)− πν]

⟩

(2)

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The cross terms have been thrown out, because they go away when the averaging isdone (since some θi are negative and some are positive). We know the average value ofcosine squared is one half. Then Eq. 2 becomes,

⟨

x2c.o.βs

⟩

=1

8 sin2 (πν)

⟨

Nerror∑

i

βiθ2i

⟩

The average of Nerror numbers is the same as Nerror times the average of the numbers.Furthermore, βi and θi are uncorrelated (the placement of errors is random with respect tothe beta function). This allows the average of the product to be written as a product ofthe averages. Then,

⟨

x2c.o.βs

⟩

=1

8 sin2 (πν)Nerror 〈βi〉

⟨

θ2i⟩

The angular deflections θi in Eq. 2 may be due to rotated dipoles, or dipoles with thewrong field strength. However, they can actually also arise when the centroid of the beampasses through a quadrupole magnet off-axis. The quadrupole will ’focus’ the entire beamin this case, rather than acting only on individual particles. The beam will get an angularkick, x

f, where x is the position of the beam centroid at the quadrupole, and f is the focal

length of the quadrupole. Considering the case of a distribution of quadrupole errors (NQ

quadrupoles in the machine), the deviation of the closed orbit due to those errors can beestimated,

σxc.o.√βs

=

√

NQ√8 sin (πν)

⟨√βi⟩

fσx

If the errors were due to rotated dipoles rather than off-axis quadrupoles, the formulawould be the same except that NQ → ND, and

σx

f→ σθ.

2 Gradient errors and correction

A gradient error (focusing strength error in a quadrupole) will shift the tune, ν, by

∆ν =β

4πf=

1

4πβB

′

L

Bρ

4

where β is the beta function at the location of the gradient error, f is the error in focalstrength, B

′

is the gradient strength, L is the length of the magnet, and Bρ = p/e is themagnetic rigidity. So, if the current is increased, the gradient B

′

goes up, and the focallength gets shorter. If the focal length in the horizontal plane were systematically shortenedaround an accelerator ring, the horizontal tune (number of horizontal betatron oscillationsper turn) would increase.

Since a quadrupole defocuses in one plane and focuses in the other, changing a quadrupolegradient will change the tune in both planes. The tune changes for a change in gradient,B

′

, from a single quadrupole located at s0 would be:

∆νx =βx(s0)

4πf=

1

4πβx(s0)

B′

L

Bρ

∆νy = −βy(s0)4πf

= − 1

4πβy(s0)

B′

L

Bρ

If it were a horizontally focusing quadrupole, an increase in gradient, B′

, would increasethe horizontal tune, but decrease the vertical tune (hence the negative sign).

Often the focusing quadrupoles in a ring are on the same circuit, and have the samestrength; while the same is true for the defocusing magnets. Changes in the gradientsof both the focusing circuit and the defocusing circuit must be made in order to haveindependent control of the horizontal and vertical tunes. The tune changes in a FODOlattice with N cells would be:

∆νx =N

4π

βHmax

(

B′

L

Bρ

)

FQ

+ βHmin

(

B′

L

Bρ

)

DQ

∆νy = −N

4π

βV min

(

B′

L

Bρ

)

FQ

+ βV max

(

B′

L

Bρ

)

DQ

3 Momentum dependent effects

The bend angle given to a particle going through a dipole magnet depends inversely on theparticle momentum, as per the relation

θ =eB0L

p

where B0 is the field in the dipole magnet, L is the length of the magnet, and e/p isthe charge to momentum ratio of the particle. Particles with momenta greater than the

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momentum of an ideal particle will travel radially outside of the design trajectory, whilethose of lower momenta will travel to the inside. The dependence of particle position on itsmomentum as it travels through a machine is described by the dispersion function, D(s).

∆x = D(s)∆p

p0

Quadrupole magnets also bend higher momenta particles less, resulting in a shorterfocal length for lower momenta particles, and longer focus for higher momenta particles.So, in an accelerator ring, the number of oscillations per turn (the tune) will be shifted foroff-momentum particles. The proportionality between the momentum spread in a beamand the resulting tune spread is called the chromaticity, ξ, of the machine.

∆ν = ξ∆p

p0

Figure 1: Particles of differing momenta entering a quadrupole lens on the same trajectorywill be bent different amounts. The lower momenta particles are bent more than highermomenta particles.

The ’natural’ chromaticity of a machine is due to the large focusing and defocusingquadrupoles in the ring (as opposed to the correction quadrupoles). An ideal particlewould have the following inverse focal length,

1

f0=eB

′

L

p0

This would shift slightly for a particle with a small momentum error, ∆p,

1

f=

1

f0+∆

(

1

f

)

=eB

′

L

p0 +∆p

=eB

′

L

p0(1 +∆pp0)=eB

′

L

p0

(

1 +∆p

p0

)

−1

6

≈ eB′

L

p0

(

1− ∆p

p0

)

where a binomial expansion was done since ∆p/p0 is small. The change in the inversefocal length must be:

∆

(

1

f

)

= −(

eB′

L

p0

)

∆p

p0

The off-momentum particle then has a tune shift,

∆ν = − β04π

(

eB′

L

p0

)

∆p

p0= − β0

4πf0

∆p

p0

A positive momentum error decreases the focusing and decreases the tune.

The chromaticity for a FODO cell can be found by summing the tune change from thetwo quadrupoles, and identifying the coefficient to ∆p/p0.

∆νx = − 1

4π

[

βHmax

fFQ

− βHmin

fDQ

]

∆p

p0

∆νy =1

4π

[

βV min

fFQ

− βV max

fDQ

]

∆p

p0

The natural chromaticities for one FODO cell are then,

ξx = − 1

4π

[

βHmax

fFQ

− βHmin

fDQ

]

ξy =1

4π

[

βV min

fFQ

− βV max

fDQ

]

4 Beam motion

Previously only the effect of quadrupole magnets was considered, giving the simplest form ofHill’s equations for horizontal and vertical motion. These were obtained using the equationsof motion;

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dx2

d2s+e

pBy = 0

dy2

d2s− e

pBx = 0

Typically in an accelerator ring there are major dipoles in the horizontal plane, and notso in the vertical. The effect of the dipoles on the motion can be found a few different ways.In the Edwards and Syphers text, cylindrical coordinates are used, so that x→ ρ0+x, where

ρ0 is the radius of curvature of the ideal trajectory. The equation of motion γm~R = e~v× ~Bis solved in cylindrical coordinates, which results in an extra term in the x equation.

x′′

+x2

ρ0= −By

Bρ

y′′

=Bx

Bρ

The extra term in the x equation represents the effect of the dipoles on particles whichare not following the design trajectory. Due to their different path lengths through thedipole, they experience more or less bend than the ideal particle. In high energy machineswith strong focusing this term is usually much smaller than the term due to strong focusingfrom the quadrupole fields. Note that also the magnetic fields have been left in generalform, so they may be expanded into a series of terms; dipole, quadrupole, sextupole, etc.The effect of each order of magnetic field in the expansion may be treated as an error thatperturbs the motion. In addition, since Bρ = p/e, the effect of momentum spread may beinvestigated.

5 Transverse resonances

Resonances that drive the amplitude of transverse beam oscillations can be understoodusing the analogy of a driven harmonic oscillator. A driving force for a harmonic oscillatoris a source of motion, and so is represented on the right side of the harmonic oscillatorequation. The driving force may be made up of sinusoidal components, and the effect ofeach component on the motion may be analyzed one by one.

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x+ ω20x = F/m

x+ ω20x = A sin (ωt+ φ) (3)

The solution for this driven harmonic oscillator is the following,

x = Asin (ωt+ φ)

ω20 − ω2

This may be verified by substitution back into the original equation. When the drivefrequency ω is equal to the natural frequency ω0, the solution for x becomes infinite. Asthe drive frequency gets close to the natural frequency ω → ω0, or ω = ω0 + δ where δ issmall compared to ω0, the oscillation of x becomes very large. Intuitively the same thingcan be expected for a particle beam if there were a drive force in the accelerator with aharmonic relation to the betatron tune. Magnetic field errors can potentially supply thisdriving force. The particles traverse the ring many times, and in this case, they will samplethe field error some significant fraction of the number of times they orbit.

Already equation 1 shows that in the presence of a dipole field error running the machineat integer tune causes particle positions to be infinite (i.e. larger than the beampipe radius).Quadrupole errors drive particle amplitudes for half integer tunes, and so on.

We’ll now check out the specific case for a sextupole error, following a discussion bySteve Holmes. Circular phase space (see emittance discussion) is convenient for analysis oferrors.

The angle φ gives the angular position of the particle in (x, βx′

+ αx) space, as shownin Fig. 2. In the absence of perturbations, φ ≡ 2πνx. Then, if the motion is stable, theamplitude A would not change from turn to turn (dA

dn= 0), and the phase evolution would

be dφdn

= 2πνx.

Consider a sextupole error that occurs as a kick in one location of the ring rather thana distributed error. Then at the location of the error, there is no position change (∆x = 0going through a thin sextupole element representing the error), but there is an angle changedue to the kick from the sextupole field. The analysis is for the x motion; so, By is thecomponent of importance since it affects the horizontal motion. To further simplify theanalysis, assume that the particle position is at y = 0 at the error location (and this won’tchange, because the perturbation is in x). Then,

θ = ∆x′

=e

p

∫

By ds =B

′′

L

2Bρx2 = sx2

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x

β(s) x’+α (s) x

r = A β(s)

φ

Figure 2: A circular trajectory in horizontal phase space of a particle at location s.

where s is a convenient symbol for the constant coefficient, and x2 = A2β cos2 φ.

Using ∆x = 0 (impulse error) the change in phase on a single turn in terms of the kickmay now be found by differentiation of x.

x = A√

β cos (φ)

∆x = 0 = ∆A√

β cos (φ)− A√

β sin (φ)∆φ

Solving for ∆φ,

∆φ = −√β

Acos (φ)∆x

′

In order to find ∆A, use Eq.4 in the emittance lecture:

A2 =x2

β+

(αx+ βx′

)2

β

the change in amplitude, ∆A, due to a kick ∆x′

can be found by differentiation. (∆x = 0)

∆A =(αx+ βx

′

)

A∆x

′

= −√

β sin (φ)∆x′

where x = A√β cos (φ), and x

′

= dxds

were used to get the expression in terms of sin (φ).We have the change in amplitude and phase of the motion due to an impulse error:

∆A = −√

β sin (φ)∆x′

∆φ = −√β

Acos (φ)∆x

′

10

To get a more specific result, use ∆x′

for a sextupole error:

dA

dn= −sA2(β)3/2 sin (φ) cos2 (φ)

dφ

dn= −sA(β)3/2 cos3 (φ) + 2πνx

where 2πνx is the normal contribution from the tune. Using trigonometric identities, thesecan be written,

dA

dn= −sA2(β)3/2

1

4(sin (φ) + cos (3φ))

dφ

dn= −sA(β)3/21

4(3 cos (φ) + cos 3(φ)) + 2πνx

Over many turns dAdn

and dφdn

average to zero, unless νx is close to integer or third integer.Sextupole errors can drive third integer resonances.

Extending the analysis to higher order errors and including skew elements, transverseresonances have the form,

mνx + nνy = p

where m, n, and p are integers of the same sign (including zero). These conditions forresonance can be plotted as lines in ’tune space’, νx versus νy. A depiction of tune space isshown in Fig. 3.

Figure 3: Plot of horizontal versus vertical tune. Solid lines are resonance lines. Thetunes νx and nuy should not lie on these lines, i.e. should not satisfy mν + nνy = p,m,n = 0, 1, 2, . . . and p = 1, 2, . . ..

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References

[1] D.A. Edwards, M.J. Syphers, ’An introduction to the physics of high energy accelera-tors’, Wiley, 1993. ISBN 0-471-55163-5

[2] Steve Holmes, Class notes for Steve Holmes USPAS course.

[3] H. Wiedemann, ’Particle Accelerator Physics I’, Springer-Verlag, 2003. ISBN 3-540-00672-9

[4] Karl L. Brown, Roger V. Servranckx, ’Optics modules for circular accelerator design’,SLAC-PUB-3957, (1986)

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